- #1
logicaljoe
- 2
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Let P(n) be the formula
1^2 + 2^2 + ... + n^2 = n(n +1)(2n +1)/6
a. Write P, is it true?(1)
1^2 + 2^2 + ... + 1^2 = 1(1 +1)(2(1) +1)/6 =
1^2 + 2^2 + ... + 1^2 = 1(1 +1)(2 +1)/6
LHS = 1^2 = 1
RHS = 2 . 3/6 = 1 True
b. Write P(k)
Basis Step
n = k
1^2 + 2^2 + ... + k^2 = k(k +1)(2k +1)/6
c. Write P(k+1)
Inductive step
n = k then n = k + 1
1^2 + 2^2 + ... + (k+1)^2 = (k+1)((k+1) +1)(2(k+1)+1)/6
Now this is where I get confused with induction writing the proof of the inductive step.
d. use mathematical induction to prove k then k+1, n>_1
1^2 + 2^2 + ... + (k+1)^2 = (k+1)((k+1) +1)(2(k+1)+1)/6
LHS = (k+1)^2
= k+1
= k
= 1
RHS = (k+3)+1(2+1)/6 combining like terms
= (k+3)+(2+1)/6
= 3k+3/6
= 6k/6
= k
= 1
Is this the kind of working that is valid? it seems right to me.
1^2 + 2^2 + ... + n^2 = n(n +1)(2n +1)/6
a. Write P, is it true?(1)
1^2 + 2^2 + ... + 1^2 = 1(1 +1)(2(1) +1)/6 =
1^2 + 2^2 + ... + 1^2 = 1(1 +1)(2 +1)/6
LHS = 1^2 = 1
RHS = 2 . 3/6 = 1 True
b. Write P(k)
Basis Step
n = k
1^2 + 2^2 + ... + k^2 = k(k +1)(2k +1)/6
c. Write P(k+1)
Inductive step
n = k then n = k + 1
1^2 + 2^2 + ... + (k+1)^2 = (k+1)((k+1) +1)(2(k+1)+1)/6
Now this is where I get confused with induction writing the proof of the inductive step.
d. use mathematical induction to prove k then k+1, n>_1
1^2 + 2^2 + ... + (k+1)^2 = (k+1)((k+1) +1)(2(k+1)+1)/6
LHS = (k+1)^2
= k+1
= k
= 1
RHS = (k+3)+1(2+1)/6 combining like terms
= (k+3)+(2+1)/6
= 3k+3/6
= 6k/6
= k
= 1
Is this the kind of working that is valid? it seems right to me.